Question

(78ab1 + 1) (78ab1 - 1) must be divisible by $2^k$, then complete set of possible values of k is (where a, b are single digit whole numbers)-

• A. {1}
• B. {1, 2}
• C. {1, 2, 3}
• D. None of these

Answer 1

Answer: (C)

{1, 2, 3}

$(78ab1+1)(78ab1-1)$

$\left(78ab2\right)\left(78ab0\right)$

If we multiply it we get the last digit as $0.$ Which is divisible $2.$

And whatever we will put the value of $a$ and $b$ we will always get the number divisible by $4$ and $8$ also$.$

So$,$ $2,2^2,2^3$ is always be the diviser of $(78ab1+1)(78ab1-1).$

Hence$,$ possible value will be $(1,2,3).$